12th NCERT INTEGRALS Exercise 7.4 Questions 25
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Integrate the functions in Exercise 1 to 23

Question (1)

\[\frac{{3{x^2}}}{{{x^6} + 1}}\]

Solution

\[I = \int {\frac{{3{x^2}}}{{{x^6} + 1}}dx} \]
We have formula
\[\int {\frac{1}{{{x^2} + {a^2}}}dx = \frac{1}{a}} \tan \left( {\frac{x}{a}} \right) + c\] We will reduce given equation to standard formula
\[I = \int {\frac{{3{x^2}}}{{{{\left( {{x^3}} \right)}^2} + {{\left( 1 \right)}^2}}}dx} \] Let x3 = t
So x6 = t2
and 3x2dx = dt
Substituting above values in given equation \[I = \int {\frac{{dt}}{{{t^2} + 1}}} \] \[I = \frac{1}{1}{\tan ^{ - 1}}\frac{t}{1} + c\] \[I = {\tan ^{ - 1}}{x^3} + c\]

Question (2)

\[\frac{1}{{\sqrt {1 + 4{x^2}} }}\]

Solution

We have the formula Here if we assume 1+4x2 = t2, then derivative will contain terms of x i.e 8xdx = 2tdt. Which is not present in numerator. For the formula to apply we have to divide equation by coefficient of x2
\[I = \frac{1}{2}\int {\frac{1}{{\sqrt {\frac{1}{4} + {x^2}} }}} dx\] \[I = \frac{1}{2}\int {\frac{1}{{\sqrt {{{\left( {\frac{1}{2}} \right)}^2} + {x^2}} }}} dx\] \[ = \frac{1}{2}\log \left| {x + \sqrt {\frac{1}{4} + {x^2}} } \right| + c'\] \[ = \frac{1}{2}\log \left| {x + \sqrt {\frac{{4{x^2} + 1}}{4}} } \right| + c'\] \[ = \frac{1}{2}\log \left| {x + \frac{{\sqrt {4{x^2} + 1} }}{2}} \right| + c'\] \[ = \frac{1}{2}\log \left| {\frac{{2x + \sqrt {4{x^2} + 1} }}{2}} \right| + c'\] \[ = \frac{1}{2}\log \left| {2x + \sqrt {4{x^2} + 1} } \right| - \frac{1}{2}\log 2 + c'\] Let c = c'- ½log2; \[ = \frac{1}{2}\log \left| {2x + \sqrt {4{x^2} + 1} } \right| + c\]

Question (3)

\[\frac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + {1^2}} }}\]

Solution

\[I = \int {\frac{1}{{\sqrt {{{\left( {2 - x} \right)}^2} + {1^2}} }}dx} \] Let 2-x = t
∴ -dx = dt
dx = -dt
Substitution gives
\[I = \int {\frac{1}{{\sqrt {{t^2} + {1^2}} }}dt} \] \[ = - \log \left| {t + \sqrt {{t^2} + 1} } \right| + c\] \[ = - \log \left| {\left( {2 - x} \right) + \sqrt {{{\left( {2 - x} \right)}^2} + 1} } \right| + c\]

Question (4)

\[\frac{1}{{\sqrt {9 - 25{x^2}} }}\]

Solution

\[I = \int {\frac{1}{{\sqrt {9 - 25{x^2}} }}dx} \] Divide by √25 \[I = \int {\frac{{\frac{1}{{\sqrt {25} }}}}{{\sqrt {\frac{9}{{25}} - {x^2}} }}dx} \] \[I = \frac{1}{5}\int {\frac{1}{{\sqrt {\frac{9}{{25}} - {x^2}} }}dx} \]
Use formula
\[\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = {{\sin }^{ - 1}}\frac{x}{a} + c} \]
\[ = \frac{1}{5}{\sin ^{ - 1}}\left( {\frac{x}{{\frac{3}{5}}}} \right) + c\] \[ = \frac{1}{5}{\sin ^{ - 1}}\left( {\frac{{5x}}{3}} \right) + c\]

Question (5)

\[\frac{{3x}}{{1 + 2{x^4}}}\]

Solution

\[I = \int {\frac{{3x}}{{1 + 2{x^4}}}dx} \] \[I = 3\int {\frac{x}{{1 + 2{{\left( {{x^2}} \right)}^2}}}} dx\]
Here in numerator there is "x" and denomenator has x4
As we assume x2 = t derivative will containterms of "x". So xdx will be replace by dt
\[I = 3\int {\frac{{\frac{{dt}}{2}}}{{1 + 2{t^2}}}} \] \[I = \frac{3}{2} \times \frac{1}{2}\int {\frac{{dt}}{{\frac{1}{2} + {t^2}}}} \] \[I = \frac{3}{4}\int {\frac{{dt}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + {t^2}}}} \]
Formula \[\int {\frac{{dx}}{{{x^2} + {a^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{x}{a}} \right)} + c\]
$$I = \frac{3}{4}\frac{1}{{\frac{1}{{\sqrt 2 }}}}{\tan ^{ - 1}}\left( {\frac{t}{{\frac{1}{{\sqrt 2 }}}}} \right) + c$$ \[I = \frac{{3\sqrt 2 }}{4}{\tan ^{ - 1}}\left( {\sqrt 2 t} \right) + c\] \[I = \frac{{3\sqrt 2 }}{4}{\tan ^{ - 1}}\left( {\sqrt 2 {x^2}} \right) + c\]

Question (6)

\[\frac{{{x^2}}}{{1 - {x^6}}}\]

Solution

\[I = \int {\frac{{{x^2}}}{{1 - {x^6}}}dx} \] \[I = \int {\frac{{{x^2}}}{{1 - {{\left( {{x^3}} \right)}^2}}}dx} \] Let x3 = t
3x2dx = dt
\[{x^2}dx = \frac{{dt}}{3}\] \[I = \int {\frac{{\frac{{dt}}{3}}}{{1 - {t^2}}}} \] \[I = \frac{1}{3}\int {\frac{1}{{1 - {t^2}}}dt} \] \[I = \frac{1}{3}\left[ {\frac{{ - 1}}{{2\left( 1 \right)}}\log \left| {\frac{{1 - t}}{{1 + t}}} \right|} \right] + c\] \[I = - \frac{1}{6}\log \left| {\frac{{1 - {x^3}}}{{1 + {x^3}}}} \right| + c\]
Use formula \[ - \log \left( {\frac{a}{b}} \right) = \log \left( {\frac{b}{a}} \right)\]
\[I = \frac{1}{6}\log \left| {\frac{{1 + {x^3}}}{{1 - {x^3}}}} \right| + c\]

Question (7)

\[\frac{{x - 1}}{{\sqrt {{x^2} - 1} }}\]

Solution

\[I = \int {\frac{{x - 1}}{{\sqrt {{x^2} - 1} }}dx} \] First we will seperate \[I = \int {\frac{x}{{\sqrt {{x^2} - 1} }}dx} - \int {\frac{1}{{\sqrt {{x^2} - 1} }}} dx\] Let x2 -1 = t2
$$\require{cancel}\cancel{2}xdx =\cancel{2}tdt $$ \[I = \int {\frac{t}{{\sqrt {{t^2}} }}dt - \int {\frac{1}{{\sqrt {{x^2} - {1^2}} }}dt} } \] \[I = t - \log \left| {x + \sqrt {{x^2} - 1} } \right| + c\] \[I = \sqrt {{x^2} - 1} - \log \left| {x + \sqrt {{x^2} - {1^2}} } \right| + c\]

Question (8)

\[\frac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}\]

Solution

\[I = \int {\frac{{{x^2}}}{{\sqrt {{x^6} + {a^6}} }}dx} \] Let x3 = t
3x2 = dt \[{x^2}dx = \frac{{dt}}{3}\] \[I = \int {\frac{{\frac{{dt}}{3}}}{{\sqrt {{t^2} + {{\left( {{a^3}} \right)}^2}} }}} \] \[I = \frac{1}{3}\int {\frac{{dt}}{{\sqrt {{t^2} + {{\left( {{a^3}} \right)}^2}} }}} \] \[I = \frac{1}{3}\log \left| {t + \sqrt {{t^2} + {a^6}} } \right| + c\] \[I = \frac{1}{3}\log \left| {{x^3} + \sqrt {{x^6} + {a^6}} } \right| + c\]

Question (9)

\[\frac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}\]

Solution

\[I = \int {\frac{{{{\sec }^2}x}}{{\sqrt {{{\tan }^2}x + 4} }}dx} \] Let tanx = t
sec2x dx = dt \[I = \int {\frac{{dt}}{{\sqrt {{t^2} + 4} }}} \] \[I = \log \left| {t + \sqrt {{t^2} + 4} } \right| + c\] \[I = \log \left| {\tan x + \sqrt {{{\tan }^2}x + 4} } \right| + c\]

Question (10)

\[\frac{1}{{\sqrt {{x^2} + 2x + 2} }}\]

Solution

\[I = \int {\frac{1}{{\sqrt {{x^2} + 2x + 2} }}dx} \]
Denominator is not given in form x2 + a2 So we will reduce it in required form x2 is first term , 2x is middle term , then last term \[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times FT}} = \frac{{{{\left( {2x} \right)}^2}}}{{4{x^2}}} = 1\] Then adjest 2 as 1+1
\[I = \int {\frac{1}{{\sqrt {{x^2} + 2x + 1 + 1} }}} dx\] \[I = \int {\frac{1}{{\sqrt {{{\left( {x + 1} \right)}^2} + {1^2}} }}} dx\] \[I = \log \left| {x + 1 + \sqrt {{{\left( {x + 1} \right)}^2} + 1} } \right| + c\] \[I = \log \left| {x + 1 + \sqrt {{x^2} + 2x + 2} } \right| + c\]

Question (11)

\[\frac{1}{{9{x^2} + 6x + 5}}\]

Solution

\[I = \int {\frac{1}{{9{x^2} + 6x + 5}}dx} \]
We dont need coefficient of x2 . So divide numerator and denominator by 9
\[I = \int {\frac{{\frac{1}{9}}}{{{x^2} + \frac{6}{9}x + \frac{5}{9}}}dx} \]
We will find Last term to make perfect square \[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times FT}} = \frac{{{{\left( {\frac{2}{3}x} \right)}^2}}}{{4{x^2}}} = \frac{{4{x^2}}}{{9 \times 4{x^2}}} = \frac{1}{9}\]
\[I = \frac{1}{9}\int {\frac{{dx}}{{{x^2} + \frac{2}{3}x + \frac{1}{9} + \frac{4}{9}}}} \] \[I = \frac{1}{9}\int {\frac{{dx}}{{{{\left( {x + \frac{1}{3}} \right)}^2} + {{\left( {\frac{2}{3}} \right)}^2}}}} \]
Use formula \[\int {\frac{1}{{{x^2} + {a^2}}}dx = \frac{1}{a}{{\tan }^{ - 1}}\frac{x}{a}} + c\]
\[I = \frac{1}{9}\left[ {\frac{1}{{\frac{2}{3}}}{{\tan }^{ - 1}}\left( {\frac{{x + \frac{1}{3}}}{{\frac{2}{3}}}} \right)} \right] + c\] \[I = \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{3x + 1}}{2}} \right) + c\]

Question (12)

\[\frac{1}{{\sqrt {7 - 6x - {x^2}} }}\]

Solution

\[I = \int {\frac{1}{{\sqrt {7 - 6x - {x^2}} }}dx} \]
adjest last term to get perfect square. Then we will get equation in a2 - x2 form \[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times \left( {FT} \right)}} = \frac{{{{\left( {6x} \right)}^2}}}{{{\rm{4}} \times {x^2}}} = \frac{{36{x^2}}}{{4{x^2}}} = 9\]
\[I = \int {\frac{1}{{\sqrt {16 - \left( {{x^2} + 6x + 9} \right)} }}} dx\] \[I = \int {\frac{1}{{\sqrt {{4^2} - {{\left( {x + 3} \right)}^2}} }}dx} \]
Use formula \[\int {\frac{1}{{\sqrt {{a^2} - {x^2}} }}dx = {{\sin }^{ - 1}}\left( {\frac{x}{a}} \right)} + c\]
\[I = {\sin ^{ - 1}}\left( {\frac{{x + 3}}{4}} \right) + c\]

Question (13)

\[\frac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}\]

Solution

\[I = \int {\frac{1}{{\sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} }}dx} \] \[I = \int {\frac{1}{{{x^2} - 3x + 2}}dx} \]
adjest last term to get perfect square. Then we will get equation in a2 - x2 form \[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times \left( {FT} \right)}} = \frac{{{{\left( {3x} \right)}^2}}}{{{\rm{4}} \times {x^2}}} = \frac{9}{4}\]
\[I = \int {\frac{1}{{\sqrt {{x^2} - 3x + \frac{9}{4} - \frac{1}{4}} }}} dx\] \[I = \int {\frac{1}{{\sqrt {{{\left( {x - \frac{3}{2}} \right)}^2} - {{\left( {\frac{1}{2}} \right)}^2}} }}} dx\] \[I = \log \left| {x - \frac{3}{2} + \sqrt {{{\left( {x - \frac{3}{2}} \right)}^2} - {{\left( {\frac{1}{2}} \right)}^2}} } \right| + c\] \[I = \log \left| {x - \frac{3}{2} + \sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} } \right| + c\]

Question (14)

\[\frac{1}{{\sqrt {8 + 3x - {x^2}} }}\]

Solution

\[I = \int {\frac{1}{{\sqrt {8 + 3x - {x^2}} }}dx} \] \[I = \int {\frac{1}{{\sqrt {8 - \left( {{x^2} - 3x} \right)} }}dx} \]
adjest last term to get perfect square. Then we will get equation in a2 - x2 form \[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times \left( {FT} \right)}} = \frac{{{{\left( { - 3x} \right)}^2}}}{{{\rm{4}} \times {x^2}}} = \frac{9}{4}\] and \[8 + \frac{9}{4} = \frac{{41}}{4}\]
\[I = \int {\frac{1}{{\sqrt {\frac{{41}}{4} - \left( {{x^2} - 3x + \frac{9}{4}} \right)} }}dx} \] \[I = \int {\frac{1}{{\sqrt {{{\left( {\frac{{\sqrt {41} }}{2}} \right)}^2} - {{\left( {x - \frac{3}{2}} \right)}^2}} }}dx} \] \[I = {\sin ^{ - 1}}\left( {\frac{{x - \frac{3}{2}}}{{\frac{{\sqrt {41} }}{2}}}} \right) + c\]

Question (15)

\[\frac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}\]

Solution

\[I = \int {\frac{1}{{\sqrt {\left( {x - a} \right)\left( {x - b} \right)} }}dx} \] \[I = \int {\frac{1}{{\sqrt {{x^2} - ax - bx + ab} }}} dx\] \[I = \int {\frac{1}{{\sqrt {{x^2} - \left( {a + b} \right)x + ab} }}} dx\]
adjest last term to get perfect square. Then we will get equation in a2 - x2 form \[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times \left( {FT} \right)}} = \frac{{{{\left( {a + b} \right)}^2}{x^2}}}{{{\rm{4}} \times {x^2}}} = \frac{{{{\left( {a + b} \right)}^2}}}{{\rm{4}}}\] and \[ab - \frac{{{{\left( {a + b} \right)}^2}}}{4} = \frac{{4ab - \left( {{a^2} + 2ab + {b^2}} \right)}}{4} = - \frac{{{{\left( {a - b} \right)}^2}}}{4}\]
\[I = \int {\frac{1}{{\sqrt {{x^2} - \left( {a + b} \right)x + \frac{{{{\left( {a + b} \right)}^2}}}{4} - \frac{{{{\left( {a - b} \right)}^2}}}{4}} }}} dx\] \[I = \int {\frac{1}{{\sqrt {{{\left( {x - \left( {\frac{{a + b}}{2}} \right)} \right)}^2} - \frac{{{{\left( {a - b} \right)}^2}}}{4}} }}} dx\] \[I = \log \left| {x - \frac{{a + b}}{2} + \sqrt {{{\left( {x - \frac{{a + b}}{2}} \right)}^2} - {{\left( {\frac{{a - b}}{2}} \right)}^2}} } \right| + c\] \[I = \log \left| {x - \frac{{a + b}}{2} + \sqrt {{{\left( {x - a} \right)}^2} - {{\left( {x - b} \right)}^2}} } \right| + c\]

Question (16)

\[\frac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}\]

Solution

\[I = \int {\frac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx} \]
Here in the numerator there is term of "x" and constant. In such case always check whether it is perfect derivative of polynomial of denominator. It is perfect then we can assume polynomical as other variable "t" and its derivative will be dt. If umerator is not proper derivative then we will find m and n such that \[num = m\frac{d}{{dx}}\left( {\text{ polynomial of denominator}} \right) + n\] In this sum numarator is proper derivative of denominator
Let 2x2 +x - 3 = t2 ⇒ t = √(2x2 +x - 3)
∴ (4x+1)dx = 2t dt \[I = \int {\frac{{4x + 1}}{{\sqrt {2{x^2} + x - 3} }}dx} \] \[I = \int {\frac{{2t}}{{\sqrt {{t^2}} }}} dt\] \[I = \int {\frac{{2\require{cancel}\cancel{t}}}{\require{cancel}\cancel{t}}dt} \] \[I = \int {2dt} \] \[I = 2t + c\] \[I = 2\sqrt {2{x^2} + x - 3} + c\]

Question (17)

\[\frac{{x + 2}}{{\sqrt {{x^2} - 1} }}\]

Solution

\[I = \int {\frac{{x + 2}}{{\sqrt {{x^2} - 1} }}dx} \]
Numerator is not perfect derivative but in derivative it has only terms of x only so we can seperate it directly
\[I = \int {\frac{x}{{\sqrt {{x^2} - 1} }}dx + \int {\frac{2}{{\sqrt {{x^2} - 1} }}} } dx\] Let x2 - 1 = t2 ⇒ t = √(x2 - 1)
$$\require{cancel}\cancel{2}xdx =\cancel{2}tdt $$ x dx = t dt \[I = \int {\frac{x}{{\sqrt {{x^2} - 1} }}dx + \int {\frac{2}{{\sqrt {{x^2} - 1} }}} } dx\] \[I = \int {\frac{t}{{\sqrt {{t^2}} }}dt + 2\int {\frac{2}{{\sqrt {{x^2} - 1} }}dx} } \] \[I = \int {dt + 2\log \left| {x + \sqrt {{x^2} - 1} } \right| + c} \] \[I = t + 2\log \left| {x + \sqrt {{x^2} - 1} } \right| + c\] \[I = \sqrt {{x^2} - 1} + 2\log \left| {x + \sqrt {{x^2} - 1} } \right| + c\]

Question (18)

\[\frac{{5x - 2}}{{1 + 2x + 3{x^2}}}\]

Solution

\[I = \int {\frac{{5x - 2}}{{1 + 2x + 3{x^2}}}dx} \]
Numerator is not a perfect derivative of denominator.
If umerator is not proper derivative then we will find m and n such that \[num = m\frac{d}{{dx}}\left( {\text{ polynomial of denominator}} \right) + n\]
\[5x - 2 = m\frac{d}{{dx}}\left( {1 + 2x + 3{x^2}} \right) + n\]
5x-2 = m(2+6x)+n
5x-2 = 2m +6mx + n
⇒ 6mx = 5x and 2m + n = -2
m = 5/6
\[2m + n = - 2\] \[\require{cancel}\cancel{2}\left( {\frac{5}{\cancel{6}3}} \right) + n = - 2\] \[n = - 2 - \frac{5}{3} = - \frac{{11}}{3}\] Now replace numerator by values of m and n \[5x - 2 = \frac{5}{6}\left( {2 + 6x} \right) - \frac{{11}}{3}\] \[I = \int {\frac{{5x - 2}}{{1 + 2x + 3{x^2}}}dx} \] \[I = \int {\frac{{\frac{5}{6}\left( {2 + 6x} \right) - \frac{{11}}{3}}}{{1 + 2x + 3{x^2}}}} \] \[I = \frac{5}{6}\int {\frac{{2 + 6x}}{{1 + 2x + 3{x^2}}}} dx - \frac{{11}}{3}\int {\frac{1}{{1 + 2x + 3{x^2}}}} dx\]
Now there are two different integration requires different method to solve. let \[{I_1} = \frac{5}{6}\int {\frac{{\left( {2 + 6x} \right)}}{{1 + 2x + 3{x^2}}}dx} \] \[{I_2} = - \frac{{11}}{3}\int {\frac{1}{{1 + 2x + 3{x^2}}}} dx\]
\[{I_1} = \frac{5}{6}\int {\frac{{\left( {2 + 6x} \right)}}{{1 + 2x + 3{x^2}}}dx} \] Let 1+2x+3x2 = t
∴ (2+6x)dx = dt \[{I_1} = \frac{5}{6}\int {\frac{{dt}}{t}} \] \[{I_1} = \frac{5}{6}\log \left| t \right| + c-1\] \[{I_1} = \frac{5}{6}\log \left| {1 + 2x + 3{x^2}} \right| + c_1\] \[{I_2} = - \frac{{11}}{3}\int {\frac{1}{{1 + 2x + 3{x^2}}}} dx\] \[{I_2} = - \frac{{11}}{9}\int {\frac{1}{{{x^2} + \frac{2}{3}x + \frac{1}{3}}}dx} \]
adjest last term to get perfect square. Then we will get equation in a2 + x2 form \[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times \left( {FT} \right)}} = \frac{{{{\left( {\frac{{2x}}{3}} \right)}^2}}}{{{\rm{4}} \times {x^2}}}\] \[LT = \frac{{\require{cancel}\cancel{4{x^2}}}}{\cancel{{36{x^2}}}9} = \frac{1}{9}\] and \[\frac{1}{3} - \frac{1}{9} = \frac{2}{9}\]
\[{I_2} = - \frac{{11}}{9}\int {\frac{1}{{{x^2} + \frac{2}{3}x + \frac{1}{9} + \frac{2}{9}}}dx} \] \[{I_2} = - \frac{{11}}{9}\int {\frac{1}{{{{\left( {x + \frac{1}{3}} \right)}^2} + {{\left( {\frac{{\sqrt 2 }}{3}} \right)}^2}}}dx} \] \[{I_2} = - \frac{{11}}{9} \times \frac{1}{{\frac{{\sqrt 2 }}{3}}}{\tan ^{ - 1}}\left( {\frac{{x + \frac{1}{3}}}{{\frac{{\sqrt 2 }}{3}}}} \right) + {c_2}\] \[{I_2} = - \frac{{11}}{{3\sqrt 2 }} \times {\tan ^{ - 1}}\left( {\frac{{3x + 1}}{{\sqrt 2 }}} \right) + {c_2}\] Now I = I1+ I2 and c = c1+ c2
\[I = \frac{5}{6}\log |1 + 2x + 3{x^2}| - \frac{{11}}{{3\sqrt 2 }} \times {\tan ^{ - 1}}\left( {\frac{{3x + 1}}{{\sqrt 2 }}} \right) + c\]
\[I = \frac{5}{6}\log |1 + 2x + 3{x^2}| - \frac{{11}}{{3\sqrt 2 }} \times {\tan ^{ - 1}}\left( {\frac{{3x + 1}}{{\sqrt 2 }}} \right) + c\]

Question (19)

\[\frac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}\]

Solution

\[I = \int {\frac{{6x + 7}}{{\sqrt {\left( {x - 5} \right)\left( {x - 4} \right)} }}dx} \] \[I = \int {\frac{{6x + 7}}{{\sqrt {{x^2} - 9x + 20} }}dx} \]
Numerator is not a perfect derivative of polynomial of derivative we will find "m" and "n" \[num = m\frac{d}{{dx}}\left( {\text{ polynomial of denominator}} \right) + n\]
\[6x + 7 = m\frac{d}{{dx}}\left( {{x^2} - 9x + 20} \right) + n\] 6x+7 = m(2x-9)+n
6x+7 = 2mx - 9m + n
⇒ 2m = 6 or m = 3
and
-9m+n = 7
substitute values o f"m"
-27+n = 7
n = 34
∴ 6x+7 = 3(2x-9) +34 \[I = \int {\frac{{6x + 7}}{{\sqrt {{x^2} - 9x + 20} }}} dx\] \[I = \int {\frac{{3\left( {2x - 9} \right) + 34}}{{\sqrt {{x^2} - 9x + 20} }}} dx\] \[I = 3\int {\frac{{\left( {2x - 9} \right)}}{{\sqrt {{x^2} - 9x + 20} }}} dx + 34\int {\frac{1}{{\sqrt {{x^2} - 9x + 20} }}} dx\]
Let I = I1 + I2 \[{I_1} = 3\int {\frac{{\left( {2x - 9} \right)}}{{\sqrt {{x^2} - 9x + 20} }}} dx\] and \[{I_2} = 34\int {\frac{1}{{\sqrt {{x^2} - 9x + 20} }}} dx\]
Solution of I1 \[{I_1} = 3\int {\frac{{\left( {2x - 9} \right)}}{{\sqrt {{x^2} - 9x + 20} }}} dx\] Let x2 -9x +20 = t
(2x-9) dx = 2tdt
Substituting values in given equation
\[{I_1} = 3\int {\frac{{2t}}{{\sqrt {{t^2}} }}dt} \] \[{I_1} = 6\int {dt} \] \[{I_1} = 6t + {c_1}\] \[{I_1} = 6\sqrt {{x^2} - 9x + 20} + {c_1}\] Solution of I2 \[{I_2} = 34\int {\frac{1}{{\sqrt {{x^2} - 9x + 20} }}} dx\]
Convert denominator in x2 - a2 form \[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times \left( {FT} \right)}} = \frac{{{{\left( {9x} \right)}^2}}}{{{\rm{4}} \times {x^2}}} = \frac{{81}}{{\rm{4}}}\] and \[20 - \frac{{81}}{4} = \frac{{ - 1}}{4}\]
\[{I_2} = 34\int {\frac{1}{{\sqrt {{x^2} - 9x + \frac{{81}}{4} - \frac{1}{4}} }}} dx\] \[{I_2} = 34\int {\frac{1}{{\sqrt {{{\left( {x - \frac{9}{2}} \right)}^2} - {{\left( {\frac{1}{2}} \right)}^2}} }}} dx\] \[{I_2} = 34\log \left| {x - \frac{9}{2} + \sqrt {{{\left( {x - \frac{9}{2}} \right)}^2} - {{\left( {\frac{1}{2}} \right)}^2}} } \right| + {c_2}\] Now I = I2 + I1 and c = c1+c2 \[I = 34\log \left| {x - \frac{9}{2} + \sqrt {{{\left( {x - \frac{9}{2}} \right)}^2} - {{\left( {\frac{1}{2}} \right)}^2}} } \right| + 6\sqrt {{x^2} - 9x + 20} + c\]

Question (20)

\[\frac{{x + 2}}{{\sqrt {4x - {x^2}} }}\]

Solution

\[I = \int {\frac{{x + 2}}{{\sqrt {4x - {x^2}} }}dx} \] Numerator is not a perfect derivative of denominator \[x + 2 = m\frac{d}{{dx}}\left( {4x - {x^2}} \right) + n\] m(4x-2x) + n = x +2
4m -2mx + n = x + 2
⇒ -2m = 1
m = -1/2
and 4m + n = 2
Substituting value of m \[\cancel{4}2\left( { - \frac{1}{\cancel{2}1}} \right) + n = 3\] -2+n = 2
n = 4 \[x + 2 = - \frac{1}{2}\left( {4 - 2x} \right) + 4\] \[I = \int {\frac{{ - \frac{1}{2}\left( {4 - 2x} \right) + 4}}{{\sqrt {4x - {x^2}} }}} dx\] \[I = - \frac{1}{2}\int {\frac{{\left( {4 - 2x} \right)}}{{\sqrt {4x - {x^2}} }}dx} + 4\int {\frac{1}{{\sqrt {4x - {x^2}} }}} dx\] Let \[{I_1} = - \frac{1}{2}\int {\frac{{\left( {4 - 2x} \right)}}{{\sqrt {4x - {x^2}} }}dx} \] let 4x - x2 = t2
(4-2x)dx = 2t dt Substitute in I1 \[{I_1} = - \frac{1}{2}\int {\frac{{2t}}{{\sqrt {{t^2}} }}dt} {\rm{ }}\] \[{I_1} = - \frac{1}{\cancel{2}}\int {\frac{{\cancel{2t}}}{\cancel{t}}dt} \] \[{I_1} = - t + {c_1}\] \[{I_1} = - \sqrt {4x - {x^2}} + {c_1}\] Now solve I2 \[{I_2} = 4\int {\frac{1}{{\sqrt {4x - {x^2}} }}} dx\] To get a2 - x2 form add and subtract 4 \[{I_2} = 4\int {\frac{1}{{\sqrt {4 - \left( {{x^2} - 4x + 4} \right)} }}} dx\] \[{I_2} = 4\int {\frac{1}{{\sqrt {{{\left( 2 \right)}^2} - {{\left( {x - 2} \right)}^2}} }}} dx\] \[{I_2} = 4{\sin ^{ - 1}}\left( {\frac{{x - 2}}{2}} \right) + {c_2}\] Now I = I1 + I2 and c = c1 + c2 \[I = - \sqrt {4x - {x^2}} + 4{\sin ^{ - 1}}\left( {\frac{{x - 2}}{2}} \right) + c\]

Question (21)

\[\frac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}\]

Solution

\[I = \int {\frac{{x + 2}}{{\sqrt {{x^2} + 2x + 3} }}dx} \]
Numerator is not a perfect derivative of polynomial of derivative we will find "m" and "n" \[num = m\frac{d}{{dx}}\left( {\text{ polynomial of denominator}} \right) + n\]
\[x + 2 = m\frac{d}{{dx}}\left( {{x^2} + 2x + 3} \right) + n\] x+2 = m(2x+2) + n
x+2 = 2mx +2m +n
⇒ 2mx = x
∴ m = 1/2
and 2m + n = 2
substituting value of m
2(1/2) + n = 2
n = 1 numarator is now \[\frac{1}{2}\left( {2x + 2} \right) + 1\] \[\therefore I = \int {\frac{{\frac{1}{2}\left( {2x + 2} \right) + 1}}{{\sqrt {{x^2} + 2x + 3} }}dx} \] \[I = \frac{1}{2}\int {\frac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}} dx + \int {\frac{1}{{\sqrt {{x^2} + 2x + 3} }}} dx\] Let I1 \[{I_1} = \frac{1}{2}\int {\frac{{2x + 2}}{{\sqrt {{x^2} + 2x + 3} }}} dx\] Let x2 + 2x +2 = t2 ⇒ t = √(x2+2x+3) taking derivative
(2x+1) dx = 2tdt \[{I_1} = \frac{1}{2}\int {\frac{{2t}}{{\sqrt {{t^2}} }}dt} \] \[{I_1} = \frac{1}{\cancel{2}}\int {\frac{{\cancel{2t}}}{\cancel{t}}dt} \] \[{I_1} = \int {dt} \] \[{I_1} = t + {c_1}\] \[{I_1} = \sqrt {{x^2} + 2x + 3} + {c_1}\] Now solve I2 \[{I_2} = \int {\frac{1}{{\sqrt {{x^2} + 2x + 3} }}} dx\]
We wil get denominator in x2 + a2 format by adjesting last term \[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times (FT)}}\] \[LT = \frac{{{{\left( {2x} \right)}^2}}}{{4{x^2}}} = 1\]
\[{I_2} = \int {\frac{1}{{\sqrt {{x^2} + 2x + 2 + 1} }}} dx\] \[{I_2} = \int {\frac{1}{{\sqrt {{{\left( {x + 1} \right)}^2} + {{\left( 1 \right)}^2}} }}} dx\] \[{I_2} = \log \left| {x + 1 + \sqrt {{{\left( {x + 1} \right)}^2} + {{\left( 1 \right)}^2}} } \right| + {c_2}\] \[{I_2} = \log \left| {x + 1 + \sqrt {{x^2} + 2x + 3} } \right| + {c_2}\] Now I = I1 + I2 and c= c1 + c2 \[I = \sqrt {{x^2} + 2x + 3} + \log \left| {x + 1 + \sqrt {{x^2} + 2x + 3} } \right| + c\]

Question (22)

\[\frac{{x + 3}}{{{x^2} - 2x - 5}}\]

Solution

\[I = \int {\frac{{x + 3}}{{{x^2} - 2x - 5}}dx} \]
Numerator is not a perfect derivative of polynomial of derivative we will find "m" and "n" \[num = m\frac{d}{{dx}}\left( {\text{ polynomial of denominator}} \right) + n\]
\[m\frac{d}{{dx}}\left( {{x^2} - 2x - 5} \right) + n = x + 3\] m(2x-2) + n = x+3
2mx - 2m +n
⇒ 2m = 1
m = 1/2
and -2m + n = 3
substituting value of m
-1 + n = 3
n = 4
∴ Numerator is x+3 = ½(2x - 2) + 4 \[I = \int {\frac{{x + 3}}{{{x^2} - 2x - 5}}dx} \] \[I = \int {\frac{{\frac{1}{2}\left( {2x - 2} \right) + 4}}{{{x^2} - 2x - 5}}dx} \] \[I = \frac{1}{2}\int {\frac{{\left( {2x - 2} \right)}}{{{x^2} - 2x - 5}}dx} + 4\int {\frac{1}{{{x^2} - 2x - 5}}} dx\] Let I1 \[{I_1} = \frac{1}{2}\int {\frac{{\left( {2x - 2} \right)}}{{{x^2} - 2x - 5}}dx} \] Let x2 -2x-5 = t
Taking drivative
(2x-2)dx = dt
Substituting values
\[{I_1} = \frac{1}{2}\int {\frac{{dt}}{t}} \] \[{I_1} = \frac{1}{2}\log \left| t \right| + c\] \[{I_1} = \frac{1}{2}\log \left| {{x^2} - 2x - 5} \right| + c_1\] Now solve I2 \[{I_2} = 4\int {\frac{1}{{{x^2} - 2x - 5}}dx} \]
We wil get denominator in x2 + a2 format by adjesting last term \[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times (FT)}}\]
\[{I_2} = 4\int {\frac{1}{{{x^2} - 2x + 1 - 6}}} dx\] \[{I_2} = 4\int {\frac{1}{{{{\left( {x - 1} \right)}^2} - {{\left( {\sqrt 6 } \right)}^2}}}} dx\] \[{I_2} = \cancel{4}2\frac{1}{{\cancel{2}1\sqrt 6 }}\log \left| {\frac{{x - 1 - \sqrt 6 }}{{x - 1 + \sqrt 6 }}} \right| + {c_2}\] Now I = I1+I2 and c = c1 + c2 \[I = \frac{1}{2}\log \left| {{x^2} - 2x - 5} \right| + \frac{2}{{2\sqrt 6 }}\log \left| {\frac{{x - 1 - \sqrt 6 }}{{x - 1 + \sqrt 6 }}} \right| + c\]

Question (23)

\[\frac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}\]

Solution

\[I = \int {\frac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}dx} \]
Numerator is not a perfect derivative of polynomial of derivative we will find "m" and "n" \[num = m\frac{d}{{dx}}\left( {\text{ polynomial of denominator}} \right) + n\]
\[m\frac{d}{{dx}}\left( {{x^2} + 4x + 10} \right) + n = 5x + 3\] ∴ m(2x+4) + n = 5x + 3
2mx + 4m + n = 5x + 3
⇒ 2m = 5
m = 5/2
and 4m + n = 3
Substitute value of m
\[\cancel{4}2\left( {\frac{5}{\cancel{2}1}} \right) + n = 3\] 10 + n = 3
n = -7
Numarator now is
\[5x + 3 = \frac{5}{2}\left( {2x + 4} \right) - 7\] \[I = \int {\frac{{5x + 3}}{{\sqrt {{x^2} + 4x + 10} }}} dx\] \[I = \int {\frac{{\frac{5}{2}\left( {2x + 4} \right) - 7}}{{\sqrt {{x^2} + 4x + 10} }}} dx\] \[I = \frac{5}{2}\int {\frac{{\left( {2x + 4} \right)}}{{\sqrt {{x^2} + 4x + 10} }}} dx - \int {\frac{7}{{\sqrt {{x^2} + 4x + 10} }}} dx\] Let I1 \[{I_1} = \frac{5}{2}\int {\frac{{\left( {2x + 4} \right)}}{{\sqrt {{x^2} + 4x + 10} }}} dx\] Let x2 + 4x + 10 = t2
⇒ (2x+4)dx = 2tdt and t = √(x2 + 4x + 10)
\[{I_1} = \frac{5}{2}\int {\frac{{2t}}{{\sqrt {{t^2}} }}} dt\] \[{I_1} = \frac{5}{\cancel{2}}\int {\frac{{\cancel{2}t}}{t}} dt\] \[{I_1} = 5\int {dt} \] \[{I_1} = 5t + {c_1}\] \[{I_1} = 5\sqrt {{x^2} + 4x + 10} + {c_1}\] Solve I2 \[{I_2} = - \int {\frac{7}{{\sqrt {{x^2} + 4x + 10} }}} dx\]
We wil get denominator in x2 + a2 format by adjesting last term \[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times (FT)}}\]
\[LT = \frac{{{{\left( {4x} \right)}^2}}}{{4{x^2}}} = \frac{{\cancel{16{x^2}}4}}{{\cancel{4{x^2}}}}\] \[{I_2} = - 7\int {\frac{1}{{\sqrt {{x^2} + 4x + 4 + 6} }}} dx\] \[{I_2} = - 7\int {\frac{1}{{\sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 6 } \right)}^2}} }}} dx\] \[{I_2} = - 7\log \left| {x + 2 + \sqrt {{x^2} + 4x + 10} } \right| + {c_2}\] Now I = I1 + I2 and c= c1 + c2 \[I = 5\sqrt {{x^2} + 4x + 10} - 7\log \left| {x + 2 + \sqrt {{x^2} + 4x + 10} } \right| + c\]

Question (24)

\[\int {\frac{{dx}}{{{x^2} + 2x + 2}}} \] equals
(A) xtan-1(x+1) +c
(B) tan-1(x+1) +c
(C) (x+1)tan-1x +c
(D) tan-1x +c

Solution

\[I = \int {\frac{{dx}}{{{x^2} + 2x + 2}}} \] Adjust the last term such that we get numerator in x2 + a2 dx \[I = \int {\frac{{dx}}{{{x^2} + 2x + 1 + 1}}} \] \[I = \int {\frac{{dx}}{{{{\left( {x + 1} \right)}^2} + {{\left( 1 \right)}^2}}}} \] \[I = \frac{1}{1}{\tan ^{ - 1}}\left( {x + 1} \right) + c\] Correct Option is (B)

Question (25)

\[\int {\frac{{dx}}{{\sqrt {9x - 4{x^2}} }}} \] equals
\[\text{(A)}\quad\frac{1}{9}{\sin ^{ - 1}}\left( {\frac{{9x - 8}}{8}} \right) + c\] \[\text{(B)}\quad\frac{1}{2}{\sin ^{ - 1}}\left( {\frac{{8x - 9}}{9}} \right) + c\] \[\text{(C)}\quad\frac{1}{3}{\sin ^{ - 1}}\left( {\frac{{9x - 8}}{8}} \right) + c\] \[\text{(D)}\quad\frac{1}{2}{\sin ^{ - 1}}\left( {\frac{{9x - 8}}{9}} \right) + c\]

Solution

We wil get denominator in a2 - x2 format by adjesting last term \[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times (FT)}}\]
\[LT = \frac{{{{\left( {9x} \right)}^2}}}{{4 \times 4{x^2}}} = \frac{{81{x^2}}}{{16{x^2}}} = \frac{{81}}{{16}}\] \[I = \int {\frac{{dx}}{{\sqrt {\frac{{81}}{{16}} - \left( {4{x^2} - 9x + \frac{{81}}{{16}}} \right)} }}} \] \[I = \int {\frac{{dx}}{{\sqrt {{{\left( {\frac{9}{4}} \right)}^2} - {{\left( {2x - \frac{9}{4}} \right)}^2}} }}} \] \[I = {\sin ^{ - 1}}\left( {\frac{{2x - \frac{9}{4}}}{{\frac{9}{4}}}} \right) \times \frac{1}{2} + c\] \[I = \frac{1}{2}{\sin ^{ - 1}}\left( {\frac{{8x - 9}}{9}} \right) + c\] Correct option is (B)
Exercise 7.3 l⇒
⇒ Exercise 7.5